用PHP显示JSON数据

我有以下数组

"forecast":{
    "txt_forecast": {
    "date":"11:00 PM EDT",
    "forecastday": [
    {
    "period":0,
    "icon":"partlycloudy",
    "icon_url":"http://icons-ak.wxug.com/i/c/k/partlycloudy.gif",
    "title":"Saturday",
    "fcttext":"Partly cloudy in the morning, then mostly cloudy. High of 88F. Winds from the South at 5 to 10 mph.",
    "fcttext_metric":"Partly cloudy in the morning, then mostly cloudy. High of 31C. Winds from the South at 10 to 15 km/h.",
    "pop":"0"
    }
    ,
    {
    "period":1,
    "icon":"partlycloudy",
    "icon_url":"http://icons-ak.wxug.com/i/c/k/partlycloudy.gif",
    "title":"Saturday Night",
    "fcttext":"Overcast in the evening, then partly cloudy. Low of 64F. Winds less than 5 mph.",
    "fcttext_metric":"Overcast in the evening, then partly cloudy. Low of 18C. Winds less than 5 km/h.",
    "pop":"10"
    }

如何在PHP中显示信息？它一直持续到第7阶段

编辑：这是我编辑过的代码，不起作用

<?php
$json_string =    file_get_contents("http://api.wunderground.com/api/7ec5f6510a4656df/geolookup/forecast/q/40121.json");
$parsed_json = json_decode($json_string);
$temp = $parsed_json->{'forecast'}->{'txt_forecast'}->{'date'};
$title = $parsed_json->{'forecast'}->{'txt_forecast'}->{'forecastday'}->{'title'};
$for = $parsed_json->{'forecast'}->{'txt_forecast'}->{'forecastday'}->{'fcttext'};
echo "Current date is ${temp},  ${title}: ${for}\n";

foreach($parsed_json['forecast']['forecastday[0]'] as $key => $value)
{
   echo $value['period'];
   echo $value['icon'];
   // etc
}
?>

更新

---

您将不得不使用该函数

$myArray = json_decode($yourJSON, true);

现在，您可以使用

foreach

从

$myArray

数组获取数据，例如：

foreach($myArray['forecast']['forecastday'] as $key => $value)
{
   echo $value['period'];
   echo $value['icon'];
   // etc
}

---

您还可以将json转换为PHP对象：

$myData = json_decode($yourJSON);

foreach($myData->forecast->forecastday as $day) {
   echo $day->period;
   echo $day->icon;
}

---

编辑： 我认为您可以通过以下方式修复问题中的代码：

<?php
$json_string =    file_get_contents("http://api.wunderground.com/api/7ec5f6510a4656df/geolookup/forecast/q/40121.json");
$parsed_json = json_decode($json_string);
$forecast = $parsed_json->forecast->txt_forecast;
$temp = $forecast->date;
$title = $forecast->forecastday->title;
$for = $forecast->forecastday->fcttext;
echo "Current date is ${temp},  ${title}: ${for}\n";

foreach($forecast->forecastday as $value) {
   echo $value['period'];
   echo $value['icon'];
   // etc
}
?>

@Eric:要将其转换为数组，这是我个人的偏好：）@DavidMorin:请参阅更新的答案plz。您没有在问题中首先提供正确的json起点：）请注意，您的json无效：您需要将其包围在

{}

中，或者删除

“预测”：

。另外，你永远不会关闭方括号。看起来你想要另一种解码形式：我的回答注意，

$parsed_json->{'forecast'}->{'txt_forecast'}->{'forecastday'}->{'fcttext'}

相当于

$parsed_json->forecast->txt_forecast->forecastday->fcttext

我想我没有理解你的意思。。你能告诉我吗？警告：为foreach（）提供的参数无效@DavidMorin:print\r（$myData->forecast->forecastday）给出了什么？@DavidMorin:print\r（$myData->forecast）？通过上面的编辑，我得到了解析错误：语法错误，意外的T\u变量in@DavidMorin：缺少分号

$myData = json_decode($yourJSON);

foreach($myData->forecast->forecastday as $day) {
   echo $day->period;
   echo $day->icon;
}

<?php
$json_string =    file_get_contents("http://api.wunderground.com/api/7ec5f6510a4656df/geolookup/forecast/q/40121.json");
$parsed_json = json_decode($json_string);
$forecast = $parsed_json->forecast->txt_forecast;
$temp = $forecast->date;
$title = $forecast->forecastday->title;
$for = $forecast->forecastday->fcttext;
echo "Current date is ${temp},  ${title}: ${for}\n";

foreach($forecast->forecastday as $value) {
   echo $value['period'];
   echo $value['icon'];
   // etc
}
?>